Calculate $\lim_{x\to\infty}\frac{x^2}{x+1}-\sqrt{x^2+1}$ I am stuck on a limit of the indeterminate form $\infty-\infty$. I have tried many approaches, such as multiplying with conjugates etc. and I am unable to find a solution. I suspect that there is an elementary trick that I am plainly missing right here. Can anybody give me a hint or solution as to solve
$$\lim_{x\to\infty}\frac{x^2}{x+1}-\sqrt{x^2+1}$$
 A: Hint: $\displaystyle\lim_{x\to\infty}\frac{x^2}{x+1}-\sqrt{x^2+1}=\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x\right)+\lim_{x\to\infty}\left(x-\sqrt{x^2+1}\right)$
A: If
$f(x)
=\dfrac{x^2}{x+1}-\sqrt{x^2+1}
$
then
$\begin{array}\\
f(x)
&=\dfrac{x^2}{x+1}-\sqrt{x^2+1}\\
&=\dfrac{x^2+x-x}{x+1}-\sqrt{x^2+1}\\
&=x-\dfrac{x}{x+1}-\sqrt{x^2+1}\\
&=x-\dfrac{x+1-1}{x+1}-\sqrt{x^2+1}\\
&=x-1+\dfrac{1}{x+1}-\sqrt{x^2+1}\\
\text{so}\\
f(x)
&\lt x-1+\dfrac{1}{x+1}-\sqrt{x^2}\\
&= -1+\dfrac{1}{x+1}\\
\text{and}\\
f(x)
&=-1+\dfrac{1}{x+1}+x-\sqrt{x^2+1}\\
&=-1+\dfrac{1}{x+1}+(x-\sqrt{x^2+1})\dfrac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\\
&=-1+\dfrac{1}{x+1}-\dfrac{1}{x+\sqrt{x^2+1}}\\
&>-1+\dfrac{1}{x+1}-\dfrac{1}{2x}\\
&=-1+\dfrac{2x-(x+1)}{2x(x+1)}\\
&=-1+\dfrac{x-1}{2x(x+1)}\\
&=-1+\dfrac{x+1-2}{2x(x+1)}\\
&=-1+\dfrac{1}{2x}-\dfrac{1}{x(x+1)}\\
\end{array}
$
so
$f(x) \to -1$.
A: I would write $$\frac{x^2-(x+1)\sqrt{x^2+1}}{x+1}=\frac{x^4-(x+1)^2(x^2+1)}{x^2(x+1)+(x+1)^2\sqrt{x^2+1}}$$
A: By binomial approximation
$$\sqrt{x^2+1}=x\left(1+\frac1{x^2}\right)^\frac12\sim x\left(1+\frac1{2x^2}\right)=x+\frac1{2x}$$
therefore
$$\frac{x^2}{x+1}-\sqrt{x^2+1}\sim\frac{x^2}{x+1}-x-\frac1{2x}=\frac{2x^3-2x^2(x+1)-1}{2x(x+1)}=\frac{-2x^2-1}{2x^2+2x}\to -1$$
A: 1) $x^2=(x+1)-1)^2=$
$(x+1)^2-2(x+1)+1$;
$\dfrac{x^2}{x+1}=(x+1)-2 +\dfrac{1}{x+1};$
2) $(x^2+1)^{1/2}=x(1+1/(x^2))^{1/2}=$
$ x +O(1/x)$;
3) $\dfrac{x^2}{x+1} -(x^2+1)^{1/2}= $
$ -1+O(1/x)$;
A: $$\frac{x^2}{x-1}-\sqrt{1+x^2}=x+1+\frac1{x-1}-\sqrt{1+x^2}=1+\frac1{x-1}-\frac1{x+\sqrt{1+x^2}}$$hence $1$.
